Three Tangent Circles – Geometry Problem With Detailed Solution
In this geometry problem, three circles are mutually tangent to each other and also tangent to a common line \( L \).
The radii of circles \( A \) and \( B \) are given, and we determine the radius of circle \( C \).
Problem
Three circles are tangent to each other and to a line \( L \).
The radius of circle \( A \) is \( 10 \text{ cm} \), and the radius of circle \( B \) is \( 8 \text{ cm} \).
Find the radius of circle \( C \).
Step-by-Step Solution
Let the radii of the three circles be:
\[
a = 10, \quad b = 8, \quad c = ?
\]
Draw perpendiculars from the centers of the circles to line \( L \).
Applying the Pythagorean theorem to the appropriate right triangles allows us to determine horizontal distances between the centers.
Step 1: Triangle \( BCB' \)
Using the Pythagorean theorem:
\[
x^2 + (b - c)^2 = (b + c)^2
\]
Expanding:
\[
x^2 + b^2 - 2bc + c^2 = b^2 + 2bc + c^2
\]
Simplifying:
\[
x^2 = 4bc
\]
\[
x = 2\sqrt{bc}
\]
Step 2: Triangle \( ACA' \)
\[
y^2 + (a - c)^2 = (a + c)^2
\]
Expanding and simplifying:
\[
y^2 = 4ac
\]
\[
y = 2\sqrt{ac}
\]
Step 3: Triangle \( ABA'' \)
\[
z^2 + (a - b)^2 = (a + b)^2
\]
Simplifying:
\[
z^2 = 4ab
\]
\[
z = 2\sqrt{ab}
\]
Step 4: Using the Geometric Relationship
Since the horizontal distance satisfies:
\[
z = x + y
\]
Substitute the expressions:
\[
2\sqrt{ab} = 2\sqrt{bc} + 2\sqrt{ac}
\]
Divide both sides by 2:
\[
\sqrt{ab} = \sqrt{bc} + \sqrt{ac}
\]
Factor \( \sqrt{c} \):
\[
\sqrt{ab} = \sqrt{c}(\sqrt{b} + \sqrt{a})
\]
Solve for \( \sqrt{c} \):
\[
\sqrt{c} = \frac{\sqrt{ab}}{\sqrt{a} + \sqrt{b}}
\]
Square both sides:
\[
c = \frac{ab}{(\sqrt{a} + \sqrt{b})^2}
\]
Step 5: Substitute the Given Values
\[
c = \frac{10 \times 8}{(\sqrt{10} + \sqrt{8})^2}
\]
\[
c \approx 2.2 \text{ cm}
\]
Final Answer
\[
\boxed{c \approx 2.2 \text{ cm}}
\]
Therefore, the radius of circle \( C \) is approximately 2.2 cm.